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I want to find critical points of $$f(x,y)=\left\{\begin{matrix} (x^2+y^2) \ln(x^2+y^2) & \text{if} &(x,y)\neq (0,0) \\ 0& \text{if} & (x,y)=(0,0) \end{matrix}\right.$$

I have started to find first derivatives:

$$f_x=2x(\ln(x^2+y^2)+1),$$ $$f_y=2y(\ln(x^2+y^2)+1).$$

Here it is my problem that I cannot find critical points here. I have tried to do like this:

$$2x(\ln(x^2+y^2)+1)=0,$$ therefore $x=0$ and $\ln(x^2+y^2)=-1$ but this one doesn't seem correct.

Parcly Taxel
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Melina
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    Try to apply polar transformation first. – Zhanxiong Aug 25 '16 at 12:57
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    As the answerers and @Zhanxiong have observed, the function is clearly constant on circles centered at the origin. Your problem thus becomes a one-dimensional one. – Lubin Aug 25 '16 at 13:58
  • There is nothing wrong with $\log(x^2+y^2)=-1$. In fact $f(x,y)$ is minimum on the circle $x^2+y^2=\mathrm{e}^{-1}$. – babakks Aug 25 '16 at 14:13

2 Answers2

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When solving problems of optimization, one should always look for transformations that simplify the problem. if you apply the coordinate trans $$ \begin{cases} x=\sqrt{r} \sin{t}\\ y=\sqrt{r} \cos{t}\\ \end{cases} $$ you end up with the function defined in $\mathbf{R}^+$ $$ g(r)=r \log{r} $$ which has got a minima for $r=\frac{1}{e}$ so the set of minima for the original function is : $$ M_n=\left\{(\frac{\sin t}{\sqrt e};\frac{\cos t}{\sqrt e})\middle| t \in [0,2\pi]\right\} $$ which is the circle of radius $\frac{1}{e}$ centered in the origin where the function f attains it's minimum value of $-\frac{1}{e}$

Remark: the origin is a local maximum as can be seen from your partial derivatives

Frank
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Use the polar form of the equation: $$f(r,\theta)=r^2\ln r^2$$ (where $f$ is zero if $r$ is zero). Since the expression only depends on $r$, $f$ is circularly symmetric about the origin.

We can then differentiate $r^2\ln r^2$ to find critical points; we find two at $r=0$ and $r=e^{-\frac12}$. Hence the critical points of $f$ in the original Cartesian formulation are the origin and the circle with equation $x^2+y^2=\frac1e$. (The origin is a removable singularity.)

Parcly Taxel
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